Let $E$ and $F$ be two real finite-dimensional vector spaces. Let $f \, : \, E \, \rightarrow \, F$ be a linear transformation from $E$ to $F$ and $H$ a vector subspace of $F$ such that $F = H + \mathrm{Im}(f)$. I would like to prove that $f^{-1}(H)$ and $H$ have the same codimension.

By definition, $\mathrm{codim}(H) = \dim(F) - \dim(H)$. Showing that $f^{-1}(H)$ has the same codimension as $H$ is equivalent to finding/constructing a subspace $G$ of $E$ such that $E = G \oplus f^{-1}(H)$ and $\dim(G) = \dim(F)-\dim(H)$. How can I construct such a subspace $G$ ?

Another idea: the Grassman formula gives : $$ \dim(F) = \dim(H) + \mathrm{rg}(f) - \dim \big( H \cap \mathrm{Im}(f) \big). $$ So, the codimension of $H$ is also equal to $\mathrm{rg}(f) - \dim\big( H \cap \mathrm{Im}(f) \big)$. So $\dim(G)$ should satisfy to: $\dim(G) + \dim\big( H \cap \mathrm{Im}(f) \big) = \mathrm{rg}(f)$.


1 Answer 1


You can proceed in the following way:

  1. The transformation $f: E \to F$ induces an injective linear transformation $$\widetilde{f}: E/f^{-1}(H) \to F/H$$
  2. The map $\widetilde{f}$ is surjective (here you need your assumption $F = H + Im(f)$).
  3. Parts 1. and 2. imply that $\widetilde{f}$ is an isomorphism, so the dimensions of domain and target are equal. Since $\dim(E/f^{-1}(H)) = \text{codim}_E(f^{-1}(H))$ and $\dim(F/H) = \text{codim}_F(H)$ you are done.
  • $\begingroup$ Thank you for your answer. I do not get why $\tilde{f}$ is injective. $\endgroup$
    – Pouteri
    Oct 2, 2017 at 13:41
  • $\begingroup$ In my opinion, it is equivalent (up to isomorphism) to consider $\tilde{f} \, : \, G \, \rightarrow \, K$ where $G,K$ are subspaces such that $E =G \oplus f^{-1}(H)$ and $F = K \oplus H$. But in that case, unless I am mistaken, $\mathrm{ker}\big( \tilde{f} \big) = \mathrm{ker}(f) \cap G$ and I do not see why $\mathrm{ker}(f) \cap G = \lbrace 0 \rbrace$. $\endgroup$
    – Pouteri
    Oct 2, 2017 at 13:50
  • $\begingroup$ You have $\widetilde{f}(x + f^{-1}(H)) = f(x) + H$ by definition and this is zero in $F/H$ if and only if $f(x) \in H$, that is $x \in f^{-1}(H)$ and so $x + f^{-1}(H)$ was zero in $E/f^{-1}(H)$ to begin with. $\endgroup$ Oct 2, 2017 at 14:22
  • 1
    $\begingroup$ Your approach also works, just note that $\ker(f) = f^{-1}(\{0\}) \subseteq f^{-1}(H)$, so $G \cap \ker(f) = \{0\}$ as desired. $\endgroup$ Oct 2, 2017 at 14:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.